Q15. Consider the first 100 natural numbers. How many of them are not divisible by any one of 2, 3, 5, 7 and 9?

We need to find the count of natural numbers from 1 to 100 that are not divisible by 2, 3, 5, 7, and 9. Since any number divisible by 9 is also divisible by 3, we only need to consider divisibility by the prime numbers 2, 3, 5, and 7.

A number that is not divisible by any of these primes (2, 3, 5, 7) must either be: 1. The number 1 (which is not divisible by any prime). 2. A prime number greater than 7. 3. A composite number whose prime factors are all greater than 7.

Let's list the prime numbers up to 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. There are 25 prime numbers in total up to 100.

Now, let's identify the numbers that fit our criteria: - The number 1: It is not divisible by 2, 3, 5, 7, or 9. (1 number) - Prime numbers greater than 7: These are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

(21 numbers) - Composite numbers whose prime factors are all greater than 7: The smallest such number would be 11 × 11 = 121, which is greater than 100. So, there are no such composite numbers within the first 100 natural numbers.

Total count = 1 (for number 1) + 21 (for primes greater than 7) = 22. This question tests basic number theory concepts like divisibility rules and prime numbers, which are fundamental to the UPSC CSAT quantitative aptitude section.